Two-Dimensional Motion and Vectors CP: 6.1

A gun with a muzzle velocity of 1000 ft/sec is shot horizontally. At the same time an identical bullet is dropped alongside the gun. Which bullet will hit first? They will land at the same time!

Assume you need to cross a river…

Say the river is 100 ft wide. 100 ft And has negligible velocity A man can kayak at 10 ft/sec. How long will it take him to cross the river?

10 seconds. Right?

100 ft Now, lets say the river is flowing at a rate of 200 ft/sec. Now how long will it take him? 10 ft/sec 200 ft/sec

100 ft Now, lets say the river is flowing at a rate of 200 ft/sec. Now how long will it take him? Still 10 seconds.

100 ft But of course he will end up 2,000 ft down stream. 10 ft/sec 200 ft/sec 2,000 ft During this 10 sec the river moves him at 200 ft/sec 2,000 ft downstream. It takes him 10 sec to cross the river

100 ft Could he make it straight across? No.

100 ft Could he make it straight across? No. 200 ft/sec

Projectile Motion Assumptions The acceleration of gravity is a constant -9.8 m/s 2 The effect of air resistance is negligible The rotation of the Earth has no effect.

Projectile motion only applies to bodies in free fall Not in free fall

Projectiles are moving in 2 dimensions Therefore, we need to look in two dimensions (the x-direction & y- direction) when solving projectile problems.

The motion on the y axis is independent of the motion on the x axis. y axis free fall motion x axis constant velocity motion.

We will see in the next chapter, this is Newton’s First Law of Motion.

On the horizontal  x = v  t On the vertical  x = v i  t+½ at 2 This leads to a parabolic path

Link Projectile Motion Applet Go to case 2

For Example… A cannon has a muzzle velocity of 62.3 m/s. What is its range when shot at an angle of o ?

1. Draw a vector diagram, and resolve the velocity vector into rectangular components m/s 30 o 62.3cos sin30 Range (  x) Example: A cannon has a muzzle velocity of 62.3 m/sec. What is its range when shot at an angle of o ?

2. Using the y axis component, and the equations of motion for free fall, calculate the time of flight. (How long the projectile is in the air) 62.3sin30 v i = 62.3sin30 = m/sec a = -9.8 m/sec 2  y = 0 t = ?  y = v i t + ½ at 2 0 = 31.15t + ½(-9.8)t 2 0 = (t)(31.15 – 4.9t) t = 6.357sec (Y axis motion only)

3. Using the time of flight, calculate how far the projectile will travel horizontally during that time.  x = v x  t  x = 62.3 cos30 x sec  x = m/sec x sec  x = ~ 343 m X Axis Motion Only

The maximum range of a projectile occurs at 45 o.

Projectiles Range of a Projectile Click here please

Misconception #1 Going faster horizontally means you don’t fall as fast.

Misconception #2: Gravity won’t act on you until you look down.

That is just so wrong!

A battleship simultaneously fires two shells at enemy ships. If the shells follow the trajectories shown, which ship gets hit first? A B 1.A will hit first3. Both will hit at the same time 2.B will hit first4. Depends on the actual angles.

A golfer makes a shot to a tee as shown. Assuming he shoots at a 60.0 o angle, with a velocity of 100. ft/sec what is the range (d x ) to the tee? (UNITS!) 60 o 75 ft R ft Example #2 Initial velocity vector

60 o 75 ft Find components of the initial velocity vector 100 cos sin 60

On the y axis a = -32 ft/sec 2 v iy = 100 sin 60 o d = + 75 ft Vertical displacement when the ball is at the elevation of the tee t = ? Using our standard equations of motion…

On the y axis a = -32 ft/sec 2 v iy = 100 sin 60 o = 86.6 d = + 75 ft t = ? d = v i t + ½ at 2 75 = (86.6)t + (-16)t 2 -16t t – 75 = 0 T = 1.08 sec. & 4.33sec As per the diagram, assume the long shot.

60 o 75ft R ft 1.08 sec 4.33 sec

On the x axis… v = 100cos60 o = 50 ft/sec t = 4.33 sec Range ( R) = v x  t = 50 ft/sec (4.33 sec) = 217 ft

Which ball spends more time in the air? Which ball has the greater launch speed? same B

The time of flight depends only on the vertical component of the initial velocity. In this case, the vertical component is the same, ie—both balls reached the same height, so they will spend the same time in the air.

Since Ball A has the shorter range, the horizontal component of its initial velocity must be less than that of Ball B. So Ball A has a smaller launching speed.

Which ball spends more time in the air? Which ball has the greater launch speed? Ball B spends more time in the air. Ball B has the greater launch speed.

Ball B spends more time in the air. Again, the time of flight depends only on the vertical component of the initial velocity. Ball B goes higher, so it must spend more time in the air.

Ball B has the greater launch speed. Both balls have the same range. We know that 45 o gives maximum range for a given speed. Equivalently, 45 o is the angle required for the smallest launch speed to achieve a given range.

Ball B has the greater launch speed. The closer the launch angle is to 45 o, the closer the launch speed is to this smallest speed. The launching angles of both balls is greater than 45 o. But, notice that Ball A’s launch angle is closer to 45 o than Ball B’s. So Ball A has the smaller launch speed of the two.